cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A067912 Engel expansion of zeta(4) = Pi^4/90 = Sum_{i>0} 1/i^4.

Original entry on oeis.org

1, 13, 15, 19, 132, 1474, 1977, 10392, 12992, 44777, 59412, 170685, 217607, 704791, 818133, 1387423, 2208674, 3206215, 12732462, 13962681, 24593168, 39744274, 55804517, 130269696, 426536424, 546807194, 1030799587, 1139987135
Offset: 1

Views

Author

Benoit Cloitre, Mar 03 2002

Keywords

Crossrefs

See A006784 for explanation of Engel expansions.

Programs

  • Mathematica
    EngelExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@
    NestList[{Ceiling[1/Expand[#[[1]] #[[2]] - 1]], Expand[#[[1]] #[[2]] - 1]/1} &, {Ceiling[1/(A - Floor[A])], (A - Floor[A])/1}, n - 1]]; EngelExp[N[Pi^4/90, 7!], 20] (* G. C. Greubel, Dec 26 2016 *)

A091833 Pierce expansion of 1/zeta(2).

Original entry on oeis.org

1, 2, 4, 7, 22, 29, 51, 173, 210, 262, 417, 746, 12341, 207220, 498538, 1286415, 2351289, 3702952, 7664494, 54693034, 75971438, 269954954, 6674693008, 13449203581, 59799655308, 98912303039, 948887634688, 3557757020909, 5898230078743
Offset: 1

Views

Author

Benoit Cloitre, Mar 09 2004

Keywords

Comments

If u(0) = exp(1/m), m integer >= 1, and u(n+1) = u(n)/frac(u(n)) then floor(u(n)) = m*n.

Crossrefs

Cf. A006784 (Pierce expansion definition), A059186.

Programs

  • Mathematica
    PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[1/Zeta[2], 7!], 25] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    default(realprecision, 100000); r=zeta(2); for(n=1, 100, s=(r/(r-floor(r))); print1(floor(r), ", "); r=s) \\ Benoit Cloitre [amended by Georg Fischer, Nov 20 2020]

Formula

Let u(0) = Pi^2/6 and u(n+1) = u(n)/frac(u(n)) where frac(x) is the fractional part of x; then a(n) = floor(u(n)).
1/zeta(2) = 1/a(1) - 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) - 1/(a(1)*a(2)*a(3)*a(4)) ...
Limit_{n->oo} a(n)^(1/n) = e.

Extensions

a(1)=1 inserted by Georg Fischer, Nov 20 2020
Showing 1-2 of 2 results.